Let $F : \mathbb{R}[X] \rightarrow \mathbb R[X]$ be a linear map and let $H \in \mathbb{R}[u,x,y,z]$ be a polynomial. Suppose that

$$ F(P \cdot Q) = H(F(P),F(Q),P,Q)$$

for all $P, Q \in \mathbb{R}[X]$. Two obvious solutions for $F$ and $H$ are

  • $F = I$, the identity function, and $H(u,x,y,z) = \alpha u x + \beta u z + \gamma yx + \delta yz$ where $\alpha,\beta,\gamma,\delta \in \mathbb{R}$ and $\alpha +\beta+\gamma+\delta = 1$;

  • $F = D$, the derivative, and $H(u,x,y,z) = u z + x y$, from the Leibniz rule.

Do there exist solutions $F$ and $H$ in which $F$ is not a linear combination of $I$ and $D$?

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    $\begingroup$ Am I the only one who did not understand the question? $\endgroup$ – Praphulla Koushik Jun 15 '19 at 14:06
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    $\begingroup$ Is H supposed to be a polynomial? $\endgroup$ – Daniil Rudenko Jun 15 '19 at 15:00
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    $\begingroup$ It might be clearer to use $\mathbb{R}[X]$ to avoid $x$ as an indeterminate in two different rings (assuming that $H$ is indeed a polynomial). $\endgroup$ – Mark Wildon Jun 15 '19 at 17:26
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    $\begingroup$ @Dattier One can make further examples by taking $F = \lambda I + \mu D$, where $F$ is the identity and $D$ is the derivative, and $H(u,x,y,z) = \lambda (\alpha u x + \beta u z + \gamma xy + \delta y z) + \mu (u z + x y)$. I think your question is interesting, but probably it won't get the attention it deserves unless you edit it to make it clearer. In particular, you might define a `non-trivial' example to be one not of the form in this comment. $\endgroup$ – Mark Wildon Jun 16 '19 at 9:49
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    $\begingroup$ @Dattier: I have rewritten the question in the usual notation, adding in the examples from my comment. You can revert it you like, but obviously I prefer my version. $\endgroup$ – Mark Wildon Jun 16 '19 at 11:45

It's not hard to see that $H$ must be of the form $\alpha ux + \beta uz + \gamma yx + \delta yz$ by $\mathbb{R}$-linearity of $F$ (see Jan-Cristoph Schlage-Puchta's answer for a fuller explanation).

By using symmetry of multiplication, we can assume without loss of generality that $\beta = \gamma$.

We have the equation $$F(P \times Q) = H(F(P), F(Q), P, Q) = \alpha F(P) F(Q) + \beta F(P) Q + \gamma P F(Q) + \delta PQ$$

Setting $P = Q = 1$, we get $F(1) = \alpha F(1)^2 + \beta F(1) + \gamma F(1) + \delta$. Degree-checking tells us that if $F(1)$ isn't constant, then $\alpha = 0$. More careful checking tells us that $\beta + \gamma = 1, \delta = 0$. This leads to the solution $F_{M, R}(P) = P \times R$, where $R = F(1)$. That answer aside, we assume $F(1)$ is constant.

I'm going to take a detour for a moment, to talk about an action of the two-dimensional nonabelian Lie group on the space of pairs $(F, H)$ that satisfy $F(P \times Q) = H(F(P), F(Q), P, Q)$. This action comes from the map $g_{(a, b)} F = aF + b \operatorname{Id}$. We want the equation to remain the same, which leads us to:

$$(gH)(u, x, y, z) = aH(\frac{u - by}{a}, \frac{x - bz}{a}, y, z) + b yz$$

$$ = \frac{\alpha}{a} ux + (\beta - \frac{b}{a} \alpha) uz + (\gamma - \frac{b}{a} \alpha) yx + (a \delta - b \beta - b \gamma + \frac{b^2}{a} \alpha + b) yz$$

Back from the detour: we now set just $Q = 1$, getting $F(P) = \alpha F(P) F(1) + \beta F(P) + \gamma P F(1) + \delta P$. Rearranging, we get $(1 - \alpha F(1) - \beta) F(P) = (\gamma F(1) + \delta) P$. The obvious solutions have $F(P) = \lambda P$ for some $\lambda \in \mathbb{R}$, which have been discussed in other answers; otherwise, we must have $1 - \alpha F(1) - \beta = \gamma F(1) + \delta = 0$.

We now split into two cases: either $\alpha = 0$, or $\alpha \neq 0$. If $\alpha = 0$, then $\beta = \gamma = 1$. Using the action of the group, we can reduce to the case that $\delta = 0$, implying that $F(P \times Q) = F(P) Q + P F(Q)$, i.e. that $F$ is a derivation. This can only happen when $F(P) = R \times \partial P$ for some polynomial $R$. Undoing the group action, we get $F_{D, R, \lambda}(P) = \lambda P + R \times \partial P$.

If $\alpha \neq 0$, then we can use the action of the group to reduce to the case $\alpha = 1, \beta = \gamma = 0$. But then $\delta = 0$, giving us the equation $F(P \times Q) = F(P) F(Q)$ - in other words, $F$ is a homomorphism. Homomorphisms from $\mathbb{R}[X]$ are compositions: $F(P) = P \circ R$ for some polynomial $R$. Undoing the group action, we get $F_{H, R, \lambda} = \lambda P + P \circ R$.

So all solutions $(F, H)$ are of the forms:

1) $F(P) = \lambda P, \frac{1}{\lambda} \alpha + \beta + \gamma + \lambda \delta = 1$

2) $F(P) = \lambda P + R \times \partial P, \alpha = 0, \beta = \gamma = 1, \delta = \lambda$

3) $F(P) = \lambda P + c (P \circ R)$ with coefficients that aren't difficult to determine using the group action.

4) $F(P) = P \times R, \alpha = 0, \beta = \gamma = \frac{1}{2}, \delta = 0$

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As there are several possibilities for $F$, here is an attempt at determining $H$.

Using the linearity of $F$ we have $H(\lambda x, y, \lambda u, v)=\lambda H(x, y, u, v)$. Taking the derivative with respect to $\lambda$ we obtain $xH_x+uH_u=H$. The map $x\partial x+u\partial u$ is linear and maps a monomial $x^ay^bu^cv^d$ to $(a+c)x^ay^bu^cv^d$, hence, every monomial with a non-zero coefficient satisfies $a+c=1$.

The same argument applies to $y$ and $v$, and we conclude that every possible polynomial $H$ is of the form $\alpha xy+\beta xv+\gamma yu + \delta uv$. On the other hand we have that if $F=\lambda\mathrm{id}$, then $H(x,y,u,v)=\alpha xy+\beta xv+\gamma yu + \delta uv$ with $\lambda\alpha+\beta+\gamma+\frac{\delta}{\lambda}=1$ is a possible polynomial $H$. In particular, all tuples $(\alpha, \beta, \gamma, \delta)$ with $\alpha\delta<0$ are possible. So further restrictions on the possible shape of $H$ are only minor.

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  • $\begingroup$ I think I'm missing something; how do you get that all tuples following the inequality are possible? It seems to me all you proved is that tuples following the equality are possible (for $F = \lambda Id$). $\endgroup$ – user44191 Jun 16 '19 at 17:50
  • $\begingroup$ @user44191: If $\alpha>0$ and $\delta<0$, then $\lambda\mapsto \lambda\alpha+\frac{\delta}{\lambda}$ is a continuous function tending to $\infty$ for $\lambda\rightarrow\infty$ and to $-\infty$ for $\lambda\rightarrow 0$. So for any given $\beta, \gamma$, there is some $\lambda$, such that $\lambda\alpha+\beta+\gamma+\frac{\delta}{\lambda}=1$. So for any such tuple there is some function $F$ (namely $\lambda\mathrm{id}$ for a suitable $\lambda$), for which this particular $H$ satisfies the functional equation. $\endgroup$ – Jan-Christoph Schlage-Puchta Jun 27 '19 at 15:52

If I understand everything correctly, another simple solution is $F(P) = P(a)$ (evaluation in a given $a\in\mathbb{R}$), with $H(u,x,y,z)=ux$.

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    $\begingroup$ I think any substitution works: for any polynomial $R$, $F_R(P) = P \circ R, H = ux$. $\endgroup$ – user44191 Jun 16 '19 at 14:00
  • $\begingroup$ @user44191 Good point. And also evaluating the derivative works. $\endgroup$ – Federico Poloni Jun 16 '19 at 14:44

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